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\usepackage{ctex} % 中文支持
\usepackage{amsmath, amssymb} % 数学公式与符号
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\usecolortheme{default} % 可选：seahorse, beaver, dolphin 等

\title{第一章：随机事件与概率}
\author{MSS ET AL}
\date{2018年9月}

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\begin{frame}{内容提要  }

\begin{itemize}

\item  随机试验、样本空间、事件

\item  事件的运算、事件域

\item  概率的公理化定义、古典方法与几何方法

\item  可加性、单调性、连续性、加法公式

\item  条件概率、全概率公式、贝叶斯公式

\item  事件之间的独立性

\end{itemize}

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\begin{frame}{本章习题}

\begin{itemize}

\item  (1.1) 1,5,7,10,11.

\item  (1.2) 1(1,2,3),3,5,10,14,23,28,29.

\item  (1.3) 1,2,4,6,10,13,17,21.

\item  (1.4) 3,5,7,8,10,15,19,23,27,33.

\item  (1.5) 1,3,7,10,13,18,22,24.

\end{itemize}

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问：什么是{\color{red}样本空间}？求下述随机试验的样本空间：

1.抛一枚硬币。
2.掷一颗骰子。
3.商场某天的顾客数。
4.某电视机的寿命。
5.测量的误差。


答：样本空间是某随机试验的所有可能发生的结果的集合。

1. $\Omega = \{ H,T \}$.\\
2. $\Omega = \{ 1,2,3,4,5,6 \}$.\\
3. $\Omega = \{ 0,1,2,3,\cdots,n,\cdots \}$.\\
4. $\Omega = \{ x\in\mathbb{R}\,\mid\, x\ge 0 \}$.\\
5. $\Omega = \{ x\in\mathbb{R} \}$.

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问：什么是{\color{red}随机事件}？投掷一颗骰子，写出下述事件所代表的集合：
1.出现1点。
2.出现偶数点。
3.出现点数大于零。
4.出现点数大于10.


答：样本空间的子集称为随机事件。\\
1. $A=\{ 1 \}$.\\
2. $A=\{ 2,4,6 \}$.\\
3. $A=\{1,2,3,4,5,6\}=\Omega$.\\
4. $A=\{x\in\Omega\,\mid\, x>10\}=\varnothing$.

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问：什么是{\color{red}随机变量}？设$X=$‘投掷一颗骰子出现的点数’，解释下述表达式的具体含义：
1.$X=3$.
2.$X>3$.
3.$X\le 6$.
4.$X=10$.


答：随机变量是样本空间到实数集的一个对应。随机变量将具体的试验结果用抽象的实数来代替。
\[X:\Omega\to\mathbb{R}.\]
\begin{align*}
\Omega{}&=\{1,2,3,4,5,6\}\\
A {}& =\{4,5,6\}= \{ \omega\in\Omega\mid X(\omega)>3\}=\{X>3\} 
\end{align*}

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问：事件之间可能有哪些关系？写出下述表达式的含义：
1.$A\subset B$.
2.$A=B$.
3.$A\cup B$.
4.$A\cap B$.
5.$AB$.
6.$A-B$.
7.$\bar{A}$.
8.$A\bigtriangleup B$.
9.$\varnothing$.
10.$\Omega$.



答：$A\subset B$ 是指若 $A$ 则 $B$ 一定发生。

事件 $A$ 发生是指集合 $A$ 中的某个样本点发生。

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问：设$A,B,C$是某随机现象的三个事件。将下述事件写成$A,B,C$的表达式：\\
1.事件D=‘A与B发生但C不发生’。\\
2.事件E=‘A，B，C中至少有一个发生’。\\
3.事件F=‘A，B，C中恰好有一个发生’。\\
4.事件G=‘A，B，C全不发生’。\\
5.事件H=‘A，B，C不全发生’。


答：1. $D=AB\bar{C}$. 
2. $E=A\cup B\cup C$.\\
3. $F=A\bar{B}\bar{C}\cup \bar{A}B\bar{C} \cup \bar{A}\bar{B}C$.

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问：设 $\Omega$ 是样本空间，设 $\cal{F}$ 为 $\Omega$ 的一些子集（即事件）组成的集合。什么样的事件组成的集合 $\cal{F}$ 称为{\color{red}事件域}？
设 $\Omega=\{\omega_1,\omega_2\}$，写出所有可能的事件域。



答：一些事件组成的集合若在事件的差集和可数并运算下封闭，则称为事件域。证明事件域在事件的可数交运算下也是封闭的。

若 $A_1,A_2,\cdots\in\mathcal{F}$ 则 $A_1-A_2\in\mathcal{F}$, $\bigcup\limits_{k=1}^{\infty} A_k\in\mathcal{F}$.

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问：一般集合的子集集合什么时候称为 $\sigma$ 域？
什么是{\color{red}博雷尔域}？证明开区间和闭区间都是博雷尔集。


答：包含半直线的最小的域称为博雷尔域。

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问：设$\Omega$是一个样本空间，设$\mathcal{F}$是$\Omega$上的一个事件域，符合哪些条件的函数$P:\mathcal{F}\to\mathbb{R}$称为一个{\color{red}概率}？


答：符合非负、正则、可数可加三条公理：
\begin{enumerate}
\item $P(A)\ge 0$.
\item $P(\Omega)=1$.
\item 若 $\forall i,j:\, A_i\cap A_j=\varnothing$, \\ 则 $P(A_1\cup A_2\cup \cdots) = P(A_1)+P(A_2)+\cdots$.
\end{enumerate}


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问：确定概率的方法有哪些？举例说明。


答：
\begin{itemize}
\item 用{\color{red}频率}去接近概率。$P(A)=\frac{n(A)}{n}$.%依据？
\item 古典方法。公式：$P(A)=\frac{|A|}{|\Omega|}$.
\item 几何方法。公式：$P(A)=\frac{S_A}{S_\Omega}$.
\end{itemize}

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问：投掷两枚硬币，求出现一正一反的概率。


答：古典方法。
\begin{align*}
\Omega {}&= \{ HH,HT,TH,TT \}\\
A {}&=\{HT,TH\}\\
P(A) {}&=\frac{|A|}{|\Omega|}=\frac{2}{4}
\end{align*}

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问：一批产品共 $N$ 件， 其中 $M$ 件是不合格的， $N-M$ 件是合格的。 随机取出 $n$ 件， 求其中有 $m$ 件不合格的概率。


答：先考虑 $m=0,m=1,m=2$ 的情形。$m=0$:
\begin{align*}
\Omega {}&= \{ (x_1,\cdots,x_n)\mid x_k=1,\cdots,N,\forall i\neq j: x_i\neq x_j \}\\
A {}&=\{(x_1,\cdots,x_n) \in \Omega \mid x_k=1,2,\cdots,N-M \}
\end{align*}

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问：甲乙两人约定在下午6-7点之间在某处见面。先到者等候20分钟。求两人能会面的概率。


答：将两人的到达时间记为$X,Y$，则会面等价于 $|X-Y|<20$.

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问：平面上画有间隔为$d$的等距平行线，向平面任意投掷长度为$\ell$的针，求针与平行线相交的概率。


答：针的位置可由两个实数确定。


问：从概率的公理化定义出发，证明：
\begin{enumerate}
\item 空集的概率是0.
\item 若有限个事件互不相容，则其和事件的概率等于其每个事件的概率的和。
\item 补事件的概率等于1减去该事件的概率。
\item 子事件的概率总是比较小。
\item 加法公式。
\end{enumerate}

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%答：
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问：
36只灯泡中有4只是60瓦的，其余是40瓦的。现任取3只，求至少取到1只60瓦的概率。


\begin{itemize}
\item 先求取到``3只都是40瓦''的概率。\\ 这是 $(32/36)(31/35)(30/34)$. 
\item 然后使用公式 $P(\bar{A})=1-P(A)$.
\end{itemize}

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问：口袋中有编号为 $1,2,\cdots,n$ 的 $n$ 个球。从中有放回地任取 $m$ 次，求取出的个球的最大号码为 $k$ 的概率。


答：先考虑$k=1,k=2$的情形。
一般情形：%$A=B-C$:\\
% \begin{align*}
% {}& \{最大号码 = k\} \\
% ={}&\{ 最大号码 \le k\} - \{最大号码 \le k-1\}
% \end{align*}

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问：已知事件 $A,B,A\cup B$ 的概率分别为 $0.4$, $0.3$, $0.6$， 求概率 $P(A\bar{B})$.


答： 画出Wenn图。

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问：在一个有$n$个人参加的晚会上，每个人带了一件各不相同的礼物。晚会期间每人从放在一起的礼物中任取一件，求至少有一人取到自己的礼物的概率。


\begin{itemize}
\item 记$A_i$=‘第$i$人取到的是自己带的礼物’。\\ 则所求事件为：$A_1\cup A_2\cup\cdots\cup A_n$.
\item 考虑公式 $P(A\cup B)=P(A)+P(B)-P(AB)$.
\end{itemize}

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问：设$(\Omega,\mathcal{F},P)$是一个概率空间，设有事件序列 
$F_1\subset F_2\subset\cdots\subset F_k\subset\cdots$,
设$F=\bigcup\limits_{k=1}^{\infty}F_k$，证明$P(F)=\lim\limits_{n\to\infty}P(F_n)$.


\begin{itemize}
\item 应用概率公理第三条：可数可加性质。
\item $F=F_1\cup(F_2-F_1)\cup (F_3-F_2)\cup\cdots$.
\end{itemize}

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问：一个家里有两个小孩。\\
%\begin{enumerate}
%\item 
(1) 求家里至少有一个女孩的概率。\\
%\item 
(2) 如果已知家里至少有一个男孩，求家里至少有一个女孩的概率。
%\end{enumerate}


答：写出样本空间和所求事件的集合。

$\Omega=\{bb,bg,gb,gg\}$,

$A='\text{至少一个女孩}'=\{bg,gb,gg\}$, 

$B='\text{至少一个男孩}'=\{bb,bg,gb\}$, 

$P(A|B)=?$

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问：写出在事件 $B$ 发生的条件下事件 $A$ 发生的{\color{red}条件概率} $P(A|B)$ 的定义。


\begin{itemize}
\item 答：$P(A|B)=\frac{P(AB)}{P(B)}$.
\item 用Wenn图解释该定义。
\item 应用：$P(AB)=P(B)P(A|B)=P(A)P(B|A)$.
\end{itemize}

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问：一批零件共100个，其中10个是次品。从中一个一个取出，求第三个才取到次品的概率。


\begin{itemize}
\item 记 $A_k=$ 第 $k$ 个取出的是正品, 则所求为 
\[ P(A_1A_2\bar{A_3}). \]
\item 应用条件概率来计算
\[ P(A_1A_2\bar{A_3}) = P(A_1)P(A_2\bar{A_3}\mid A_1) = \]
\end{itemize}

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问：设罐中有$b$个黑球，$r$个红球。每次随机取出一球然后又放回，还加进$c$个同色球和$d$个异色球。记$B_i$为‘第$i$次取出的是黑球’，$R_j$为‘第$j$次取出的是红球’。
计算概率$P(B_1R_2R_3)$.

%不放回抽样，放回抽样，传染病模型，安全模型。



答：应用条件概率来计算。
\begin{align*}
 P(B_1R_2R_3) {}&= P(B_1) P( R_2R_3 \mid B_1)\\
 {}&= P(B_1) P(R_2\mid B_1) P(R_3\mid R_2B_1)
\end{align*}

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问：用全概率公式计算：在$n$张彩票中有一张是奖券，求第二人摸到奖券的概率。


\begin{itemize}
\item 以第一人是否摸到奖券作为分类条件，写出全概率公式。
\item 设 $A_k =$ ‘第 $k$ 人摸到奖券’， 则 
\[ P(A_2) = P(A_1)P(A_2\mid A_1) + P(\bar{A_1})P(A_2\mid \bar{A_1}) \]
\end{itemize}

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问：用全概率公式计算：易出事故者1年内发生事故的概率为0.4, 占比例20\%; 不易出事故者1年内发生事故的概率为0.1, 占比例80\%. 某人来投保，求其一年内出事故的概率。


\begin{itemize}
\item 以某人是否为‘易出事故者’作为分类条件。%写出全概率公式。
\item $A=$‘某人为易出事故者’，\\ $B=$‘其一年内出事故’。
\end{itemize}

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问：某病发病率为0.0004，已知有病者化验结果正确率99\%, 无病者化验结果正确率99.9\%. 现某人化验结果认为有病，求其真的有病的概率。


%答：用{\color{red}贝叶斯公式}计算。
\begin{itemize}
\item 以某人是否真的有病作为分类条件。%写出全概率公式。
\item $A=$ 某人化验结果认为有病，\\ $B=$ 某人真的有病。
\item $P(B\mid A)=\frac{P(BA)}{P(A)}$ 这里是{\color{red}贝叶斯公式}。
\end{itemize}

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问：称事件 $A$ 与事件 $B$ {\color{red}独立}，是指什么？\\
称三个事件 $A,B,C$ 相互独立，又是指什么？


\begin{itemize}
\item $P(AB)=P(A)P(B)$.
\item $P(ABC)=P(A)P(B)P(C)$,\\ $P(AB)=P(A)P(B)$, \\ $P(AC)=P(A)P(C)$, \\ $P(BC)=P(B)P(C)$.
\end{itemize}

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问：从一副52张扑克牌中任取1张。记事件 $A$ =‘取到黑桃’, $B$ =‘取到J’. 
判断事件$A, B$是否独立。


\begin{itemize}
\item 要计算概率来判断，不能用直觉来判断。
\item 计算 $P(AB)$ 与 $P(A)P(B)$, 比较是否相等。
\item 如果52换成54，结果一样吗？
\end{itemize}

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问：已知家里有3个小孩。记事件 $A$ =‘男孩女孩都有’，$B$ =‘至多一个女孩’。判断事件 $A, B$ 是否独立。


\begin{itemize}
\item 计算 $P(AB)$ 与 $P(A)P(B)$, 比较是否相等。
\item 样本空间 $\Omega=\{bbb,bbg,bgb,gbb,\cdots\}$
\item $P(A)=6/8$, $P(B)=1/2$, $P(AB)=3/8$.
\end{itemize}

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问：证明：若事件 $A, B, C$ 相互独立，则 $A-B$ 与 $C$ 独立。$AB$ 也与 $C$ 独立。


\begin{itemize}
\item 验证等式 $P[(A-B)C]=P(A-B)P(C)$.
\item 验证等式 $P[(AB)C]=P(AB)P(C)$.
\end{itemize}

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问：两射手独立地向同一目标射击。击中的概率分别为0.8和0.9. 求至少一人击中目标的概率。


\begin{itemize}
\item 设 $A=$‘甲击中目标’, $B=$‘乙击中目标’
\item $P(A\cup B)=P(A)+P(B)-P(AB)$
\item 由独立性可得 $P(AB)=P(A)P(B)$
\end{itemize}

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问：某零件用第一种工艺加工，三道工序出错概率分别是0.3,0.2,0.1; 用第二种工艺加工，二道工序出错概率分别是0.3,0.2. 求哪种工艺比较可靠?


\begin{itemize}
\item 工艺一成功概率 = $(0.7)(0.8)(0.9)=0.504$
\item 工艺二成功概率 = $(0.7)(0.8)=0.56$
\end{itemize}

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问：两名选手轮流对同一目标射击。甲命中概率为$\alpha$, 乙命中概率为$\beta$. 设甲先射击。求甲先命中的概 率。


答：甲先命中的概率为
\begin{align*}
{}& \alpha+(1-\alpha)(1-\beta)\alpha+(1-\alpha)^2(1-\beta)^2\alpha+\cdots \\
={}&\alpha \frac{1}{1-(1-\alpha)(1-\beta)}
=\frac{\alpha}{\alpha+\beta-\alpha\beta}
\end{align*}

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问：设每个元件正常工作的概率为0.9. 分别求串联、并联、桥式系统正常工作的概率。


答：

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问：两个试验称为是独立的，是指什么？


答：
定义：如果试验E1的任何结果与试验E2的任何结果都独立，则称这两试验独立。

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问：设彩票每次中奖概率为10万分之一。设每次开奖都是独立的。求520次都未中奖的概率。


答：

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